In the Krylov-Bogoliubov theorem, the transformation is assumed continuous. Does the theorem hold if the transformation is assumed only to be measurable? If not, what is a counterexample?

Edit A few proofs of the Krylov-Bogoliubov theorem are given here, but I'm not sure if the proofs carry through with the transformation just measurable.


No, it does not hold for merely measurable transformations. As a counterexample, let the space be the nonnegative integers $X=\mathbb N$ with the sigma-algebra $\mathcal{P}(\mathbb N)$ (the power set). Let $F\colon X\to X$ be given by $F(x)=x+1$. If $\mu$ is a finite $F$-invariant measure then every $A\in\mathcal{B}$ must have measure 0.

For any nonnegative integer $n$, $F$-invariance gives $\mu([n,\infty))=\mu(F^{-n}([n,\infty)))=\mu(\mathbb{N})$. So, by monotone convergence $$ \mu(\mathbb{N})=\lim_{n\to\infty}\mu([n,\infty))=\mu(\emptyset)=0. $$

See also the following questions and answers, Probability of picking a random natural number and Why isn't there a uniform probability distribution over the positive real numbers?.

Just in case you are concerned that this example is not a compact metrizable space, consider the metric on $\mathbb{N}$ given by $d(x,y)=\vert\theta(x)-\theta(y)\vert$, where $\theta(x)=\frac1x$ for $x\not=0$ and $\theta(0)=0$. This is a metric making $\mathbb N$ into a compact space, with Borel sigma-algebra equal to the power set.

  • $\begingroup$ Thanks. I was surprised that the $F$ you gave isn't continuous. $\endgroup$ – Quinn Culver Jan 4 '12 at 20:14
  • $\begingroup$ To show that every set must have measure $0$, you can also notice that $\mu(\{n+1\}) = \mu(F^{-1}\{n+1\}) = \mu(\{n\})$ for each $n$, so every singleton is null. Since the space is countable, the result follows. $\endgroup$ – Quinn Culver Jan 4 '12 at 20:15

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